Question

Sql Query：

SELECT test.* FROM test JOIN test_shares
WHERE test.test_id = test_shares.test_id 
AND test_shares.email_address = '$email' 
AND test.url is NOT NULL ORDER BY title ASC;

重写为Zend_Db_Select声明。

$select = $db->select ()
    ->from ( 'test', 'test.*' )
    ->join ( 'test_shares', array () )
    ->where ( 'test.test_id = test_shares.test_id')
    ->where ( 'test_shares.email_address = ?',  '$email')
    ->where ( 'test.url is NOT NULL')
    ->order ( 'title' );
echo $select;

输出：

SELECT test.* FROM test JOIN test_shares ON Array 
WHERE test.test_id = test_shares.test_id 
AND test_shares.email_address = '$email' 
AND test.url is NOT NULL ORDER BY title ASC

请说明为什么会出现'ON Array'。

Answer 1

试试这个：

$select = $db   ->select ()
    ->from ( 'movie', 'movie.*' )
    ->join(array('movie_shares'),  'movie.movie_id = movie_shares.movie_id');    
    ->where ( 'movie_shares.email_address = ?',  '$email')
    ->where ( 'movie.url is NOT NULL')
    ->order ( 'title' );
echo $select;

您正在提供一个数组，其中ON子句需要一个字符串，因此PHP实际上打印array()而不是字符串。

当我回显Zend_Db查询时，它包含'ON Array'，为什么？

1 个答案: